why malloc() fail!

>> you cannot realloc a memory block that has not been malloced.
Partially true.  You can realloc() a NULL pointer.  Works just like malloc().

I'v  malloced it !
but it not work .Pelease Help me!
 

How did you malloc it?  How do you call CreateMsg?

please show the call to CreateMsg including the code for malloc

check to see if the char* pointer whose address you pass to CreateMsg hasn't alreday been freed.  Or that you aren't passing a NULL (instead of pointer to NULL).  I'd include a check that MsgData != NULL and *MsgData != NULL at the start of your function.

please show the call to CreateMsg including the code for malloc

check to see if the char* pointer whose address you pass to CreateMsg hasn't alreday been freed.  Or that you aren't passing a NULL (instead of pointer to NULL).  I'd include a check that MsgData != NULL and *MsgData != NULL at the start of your function.

My program is:

    int CreateMsg(MsgData)
    char **MsgData;
    {
    short int MsgLen,RefID,MsgType;
    int RegNum=0;
    int Len,i,CurLen=0;
    char tmp;

    MsgType=0x0107;
    MsgLen=100;

    RefID=0x0501;

    *MsgData=(char *)realloc(*MsgData,MsgLen);
    if(!*MsgData)
    {
    fprintf(stdout,"\nOut of memory \n");
    return(0);
    }
             .
             .
             .

    return(MsgLen);
    }


Tthe call to CreateMsg is :

Msg=(char *)malloc(sizeof(char));
if(!Msg)
{
      printf("\nOut of memory");
      return(1);
}

MsgLen=CreateMsg(&Msg);

I need the CreateMsg return ' MsgLen' and the created 'Msg'  same time!
 

> Msg=(char *)malloc(sizeof(char));

your sizeof(char) should bee sizeof(char*), because
*Msg is of type (char*), and not char.

Your problem is that you are trying to assign a pointer (normally 4 bytes long) to an allocated space of only 1 byte.

I have copied your code, added two statements, compiled
and run with MS VC++ on a PC running Windows NT and it
was just fine.  The code is enclosed below.  Are you sure
the core dumped happened at the realloc()?
========================================================
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <string.h>

int CreateMsg( char **MsgData )
{
    short int MsgLen, RefID, MsgType;
    int RegNum=0;
    int CurLen=0;

    MsgType = 0x0107;
    MsgLen = 100;
    RefID = 0x0501;

    *MsgData = ( char * )realloc( *MsgData, MsgLen );

    if( ! *MsgData )
    {
        fprintf( stdout, "\nOut of memory\n" );
        return( 0 );
    }

    return( MsgLen );
}

main()
{
    char *Msg;
    int MsgLen;

    Msg = ( char * )malloc( sizeof( char ));

    if( ! Msg )
    {
        printf( "Out of memory\n" );
        return( 1 );
    }

    MsgLen = CreateMsg( & Msg );
    strcpy( Msg, "hello world" );
    printf( "%s\n", Msg );
}
==========================================

Good luck.

sizeof(char) was fine.  Please ignore the comment above about sizeof(char) needing to be sizeof(char*).

Thank very much sir!  I'm so sorry  for my mistake that has bother you so much!
Last nigth ,I find the  error in the 'Msg' declaration ,which is :
char            Msg+NULL;

aha .. you declared Msg as a 'char' instead of a 'cahr*' .. is that correct?  It is odd that the compiler did not generate a warning error for this.  If your compiler has options for warning levels you should ALWAYS compile your code with the maximum level of warnings.  Such a mistake should have been picked up.

Teach me about the compiler 's  options for warning levels please!
Thanks!!!

I don't know which compiler you have ... check your compiler's documentation.

If it is Visual C++ then I _can_ help (because that is what I work with)

Thanks!